【数学A】例題1.2.21：重複組合せ（One More）★★★

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% 例題A1.2.21：重複組合せ（One More）★★★
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a，b，cの3個の文字の中から，重複を許して5個取り出す組合せは何通りあるか．

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% 例題A1.2.21の解答（One More）★★★
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取り出す5個の文字を$\bigcirc$で表し，3種類の文字の区切りを2本の$|$で表すとする．

5個の$\bigcirc$と2本の$|$を1列に並べて，

1本目の$|$より左側にある$\bigcirc$はすべてa，

1本目と2本目の$|$の間にある$\bigcirc$はすべてb，

2本目の$|$より右側にある$\bigcirc$はすべてc

を表すとする．

このとき，a，b，cから重複を許して5個取り出す組合せは，5個の$\bigcirc$と2本の$|$を並べる順列に一致する．

よって，求める組合せの総数は，

\[
\frac{7!}{5!2!}=21(\text{通り})
\]

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% 例題A1.2.21の別解
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$\bigcirc$と$|$の個数を合わせた$5+2=7$（個）の場所から，$\bigcirc$が入る5個の場所を選ぶと考えられるから，

\[
{}_7\mathrm{C}_5={}_7\mathrm{C}_2=21(\text{通り})
\]

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% 問題A1.2.21：重複組合せ（One More）★★★
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x，y，z，wの4個の文字の中から，重複を許して6個取り出す組合せは何通りあるか．

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% 問題A1.2.21の解答（One More）★★★
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取り出す6個の文字を$\bigcirc$で表し，4種類の文字の区切りを3本の$|$で表すとする．

6個の$\bigcirc$と3本の$|$を1列に並べて，

1本目の$|$より左側にある$\bigcirc$はすべてx，

1本目と2本目の$|$の間にある$\bigcirc$はすべてy，

2本目と3本目の$|$の間にある$\bigcirc$はすべてz，

3本目の$|$より右側にある$\bigcirc$はすべてw

を表すとする．

このとき，x，y，z，wから重複を許して6個取り出す組合せは，6個の$\bigcirc$と3本の$|$を並べる順列に一致する．

よって，求める組合せの総数は，

\[
\frac{9!}{6!3!}=84(\text{通り})
\]

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% 問題A1.2.21の別解
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$\bigcirc$と$|$の個数を合わせた$6+3=9$（個）の場所から，$\bigcirc$が入る6個の場所を選ぶと考えられるから，

\[
{}_9\mathrm{C}_6={}_9\mathrm{C}_3=84(\text{通り})
\]

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% 例題A1.2.21の側注：重複組合せ（One More）★★★
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% 側注1
3種類の文字を表すには，$3-1=2$（本）の$|$を用いる．

% 側注2
対応例（aabbc）
\[
\underbrace{\bigcirc\bigcirc}_{\text{a（2個）}}|\underbrace{\bigcirc\bigcirc}_{\text{b（2個）}}|\underbrace{\bigcirc}_{\text{c（1個）}}
\]

% 側注3
3種類から5個取り出す重複組合せ${}_3\mathrm{H}_5={}_{3+5-1}\mathrm{C}_5$

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% 問題A1.2.21の側注：重複組合せ（One More）★★★
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% 側注1
4種類の文字を表すには，$4-1=3$（本）の$|$を用いる．

% 側注2
4種類から6個取り出す重複組合せ${}_4\mathrm{H}_6={}_{4+6-1}\mathrm{C}_6$

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% 例題A1.2.21の考え方・補足など：重複組合せ（One More）★★★
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% 考え方
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3種類から5個取り出す重複組合せである（下の余談を参照）．例題では，取り出した組合せ（順列ではない）を考えており，重複順列$3^5$とは異なるので注意すること．

5個の$\bigcirc$を2本の仕切り$|$で3つに分けると考えると，5個の$\bigcirc$と2個の$|$の合計7個の同じものを含む並び方の総数を求めればよい．このとき，$\bigcirc$，$|$と文字a，b，cは右のように対応する．

[width=6cm]

\begin{align*}
&\text{aaaaa}&\cdots&\bigcirc\bigcirc\bigcirc\bigcirc\bigcirc||\\
&\text{aabbc}&\cdots&\bigcirc\bigcirc|\bigcirc\bigcirc|\bigcirc\\
&\text{aaccc}&\cdots&\bigcirc\bigcirc||\bigcirc\bigcirc\bigcirc\\
&\text{bbbbb}&\cdots&|\bigcirc\bigcirc\bigcirc\bigcirc\bigcirc|\\
&\text{bcccc}&\cdots&|\bigcirc|\bigcirc\bigcirc\bigcirc\bigcirc
\end{align*}

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% 余談
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一般に，$n$種類のものから重複を許して$r$個を選ぶ組合せを重複組合せといい，総数を${}_n\mathrm{H}_r$で表す（Hは同次積homogeneous productの頭文字である）．

これは，$r$個の$\bigcirc$と，それを$n$種類に分けるための$(n-1)$本の仕切り|を合わせた合計$(n+r-1)$個のものの並び方の総数と等しい．よって，$(n+r-1)$個の場所の中から$\bigcirc$に入る$r$個の場所を選ぶ組合せの数，または，$r$個の$\bigcirc$と$(n-1)$本の仕切りを含む順列の総数として考えることができるから，

\[
{}_n\mathrm{H}_r={}_{n+r-1}\mathrm{C}_r=\frac{(n+r-1)!}{r!(n-1)!}
\]